# Chapter 6:Slope and Deflection¶

## Problem no 6.1,Page No.154¶

In :
import math

#Initilization of variables

b=0.12 #m #Width of beam
d=0.2  #m #Depth of beam
dell=0.005 #m #Deflection
E=2*10**5*10**6 #N/m**2
L=2.5 #m #Length of beam

#Calculations

I=b*d**3*12**-1 #m**4 #M.I of rectangular section
w=8*E*I*dell*(L**4)**-1 #N/m #U.d.l

#Let slope at free end be theta

theta_2=W*L**2*(2*E*I)**-1 #Slope at free end

#Result
print"Uniformly distributed Load beam should carry is",round(w,2),"N/m"
print"Concentrated Load at free end is",round(W,2),"KN"

Uniformly distributed Load beam should carry is 16384.0 N/m
Concentrated Load at free end is 15.36 KN


## Problem no 6.2,Page No.155¶

In :
import math

#Initilization of variables

L=6 #m #Length of beam
y_b=1.5*10**-2 #m #Deflection
E=2*10**7*10**4
sigma=10*10**3*10**4
#d=2*b

#Calculations

#Let w*I**-1=X    #From Deflection at the free end Equation
X=y_b*8*E*(L**4)**-1*10**-3    #Equation 1

#Let w*b*I**-1=Y     #From Max bending stress at the extreme fibre From N.A
Y=sigma*2*(L**2)**-1           #Equation 2

b=Y*X**-1  #width of beam #mm
d=2*b      #depth of beam #mm

#Result
print"The Dimension of Beam are:b",round(b,2),"mm (width)"
print"                         :d",round(d,2),"mm (depth)"

The Dimension of Beam are:b 300.0 mm (width)
:d 600.0 mm (depth)


## Problem no 6.3,Page No.156¶

In :
import math

#Initilization of variables

L=3 #m #Length of beam
L_1=1.2 #m #Distance from fixed end
d=0.25 #m #Depth of beam
w=15*10**3 #N #U.d.L
E=2*10*10**4 #N/m**2
I=13500*10**-4 #M.I

#Calculations

y_b=W*L_1**3*(3*E*I)**-1+W*L_1**2*(2*E*I)**-1*(L-L_1)+w*L**4*(8*E*I)**-1 #Deflection at free end

M=W*L_1+w*L*L*2**-1 #Max Bending moment at the fixed end A #Nm
y=d*2**-1
sigma_max=M*y*I**-1  #N/cm**2 #Max Bending stress at extreme fibre

#Result
print"Deflection at the free end is",round(y_b,4),"cm"
print"Max stress due to bending is",round(sigma_max,2),"N/cm**2"

Deflection at the free end is 0.8398 cm
Max stress due to bending is 10694.44 N/cm**2


## Problem no 6.4,Page No.156¶

In :
import math

#Initilization of variables

M=100*10**3 #N #Moment
L=3 #m #Length
d=0.15 #m #Width
b=0.1 #m #width
E=2.1*10**7*10**4 #N/cm**2

#Calculations

I=b*d**3*12**-1 #cm**4 #M.I of beam section
B_2=M*L**2*(2*E*I)**-1*10**2 #cm #Deflection at point B

#Result
print"The slope at Point B is",round(B_1,2),"radian"
print"The Deflection at point B is",round(B_2,2),"cm"

The slope at Point B is 0.05 radian
The Deflection at point B is 7.62 cm


## Problem no 6.5,Page No.157¶

In :
import math

#Initilization of variables

b=0.1 #m #width
d=0.2 #m #depth
L=2 #m #Length of beam
L_1=1 #m #Length from free end
E=210*10**9
w=2*10**3 #N/m

#Calculations

I=b*d**3*12**-1 #m**4 #M.I of the beam section

#Slope at free end
theta=W*L**2*(2*E*I)**-1+w*L**3*(6*E*I)**-1-w*(L-L_1)**3*(6*E*I)**-1

#Deflection at free end
y_b=(W*L**3*(3*E*I)**-1+w*L**4*(8*E*I)**-1-w*(L-L_1)**4*(8*E*I)**-1-w*(L-L_1)**3*L_1*(6*E*I)**-1)*10**3

#Result
print"Deflection at free end is",round(y_b,2),"mm"

Slope at free end is 0.00031 radian
Deflection at free end is 0.43 mm


## Problem no 6.6,Page No.158¶

In :
import math

#Initilization of variables

L=10 #m #span of beam
a=6 #m #Distance from left end of beam to point Load
b=4 #m ##Distance from right end of beam to point Load
E=210*10**9
I=10**-4 #m #M.I of beam

#Calculation

#slope at left end is given by

y_c=W*a*b*(L**2-a**2-b**2)*(6*E*I*L)**-1*10**3 #m

#Maximum Deflection of the beam is
y_max=W*a*(L**2-a**2)**1.5*(15.588457*E*I*L)**-1*10**3 #m

#Result
print"#Maximum Deflection of the beam is",round(y_max,2),"m"

slope at left end is 0.00267 radian
Deflection under Load is 9.14 m
#Maximum Deflection of the beam is 9.38 m


## Problem no 6.7,Page No.158¶

In :
import math

#Initilization of variables

L=5 #m #Length of beam
w=40*10**3 #N #U.d.L
y_max=0.01 #Deflection
sigma_s=7*10**6 #Bending stress
E=10.5*10**9

#Calculation

M=w*L*8**-1 #N*m #Max Bending moment

#From equation of max deflection
I=5*w*L**3*(y_max*384*E)**-1 #m**4

d=sigma_s*2*I*M**-1*10**2 #cm

#Result
print"Minimum value of Depth is",round(d,2),"cm"

Minimum value of breadth is 17.77 cm
Minimum value of Depth is 34.72 cm


## Problem no 6.8,Page No.159¶

In :
import math

#Initilization of variables

L=6 #m #Length of beam
d=0.15 #m #diameter
y_max=1.035*10**-2 #m #Deflection
E=210*10**9

#Calculations

I=pi*64**-1*d**4 #M.I of Beam
theta_A=3*y_max*L**-1
theta_B=-theta_A

#Result
print"The Heaviest central Point Load placed is",round(W,2),"N"

The Heaviest central Point Load placed is 483000000.0 N
Slope at supports are:theta_A 0.00517 radian


## Problem no 6.9,Page No.160¶

In :
import math

#Initilization of variables

L=14 #m #Lenth of steel girder
E=210*10**9 #modulus of Elasticity of steel
I=16*10**4*10**-8 #M.I of girder section

#Calculations

#R_a+R_b=200  #R_a & R_b are the Reactions at supports A & B respectively

#After taking moment at B We get
R_a=(120*11+80*4.5)*14**-1 #KN
R_b=200-R_a

#After considering section at X-X at a distance x from left end A and taking B.M at X-X
#M=120*x-120(x-3)-80*(x-9.5)

#After Integrating twice we get
#EI*dy*dx**-1=-60*x**2*+60(x-3)**2+40(x-9.5)**2+C_1 #slope

#Again on Integrating we get
#EI*y=-20*x**3+20(x-3)**3+40*3**-1*(x-9.5)**3+C_1*x+C_2 #Deflection

#At A deflection is zero,i.e at x=0,y=0
#At B deflection is zero,i.e at x=14,y=0 So C_2=0

C_1=-(-20*(14)**3+20*(11)**3+40*3**-1*(14-9.5)**3)*14**-1 #constant

#Now Deflection at D i.e at x=3 m
x=3
y_D=1*(E*I)**-1*(-20*x**3+20*(x-3)**3+C_1*x)*10**3

#Now Deflection at D i.e at x=9.5 m
x=9.5
y_C=1*(E*I)**-1*(-20*x**3+20*(x-3)**3+40*3**-1*(x-9.5)**3+C_1*x)*10**3

#Result
print"Deflection under points of two Loads are i.e: at pt D",round(y_D,4),"m"
print"                                            : at pt C",round(y_C,4),"m"

Deflection under points of two Loads are i.e: at pt D 0.0156 m
: at pt C 0.0199 m


## Problem no 6.10,Page No.161¶

In :
import math

#Initilization of variables

E=200*10**9 #Pa
I=20000*10**-8 #m**4

#Calculations

#Now Taking moment at B
R_a=(1000*3*4.5+1000*2)*6**-1 #Reaction Force at pt A

#On part BC u.d.l of 1KN/m is introduced both above and below
#consider section at distance x i.e X-X and considering moment at section X-X

#M=15500*x*6**-1-1000*x**2*2**-1-1000(x-4)+1000*2**-1*(x-3)**2
#EI*d**2y*d**x=-M=15500*x*6**-1-1000*x**2*2**-1-1000(x-4)+1000*2**-1*(x-3)**2

#Now Integrating above Equation we get Equation of slope
#EI*dy*dx**-1=-15500*x**2*12**-1+1000*x**3*6**-1+1000*(x-4)**2*2**-1+1000*6**-1*(x-3)**3+C_1

#Now Integrating above Equation we get Equation of Deflection
#EI*y=-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x+C_2

#At x=0,deflection is zero,i.e y=0  C_2=0
#At x=6,deflection is zero,i.e y=0
x=6
C_1=-(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*(x-3)**4*24**-1)*x**-1 #Constant

#Answer for constant C_1 is incorrect in Book

#Now Deflection at C,put x=3 m
x=3
y_C=1*(E*I)**-1*(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x)*10**3

#Now Deflection at D,put x=4 m
x=4
y_D=1*(E*I)**-1*(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x)*10**3

#Answers for y_C & y_D are incorrect in book

#Result
print"Deflection at pt C is",round(y_C,2),"mm"
print"Deflection at pt D is",round(y_D,2),"mm"

Deflection at pt C is 0.22 mm
Deflection at pt D is 0.15 mm


## Problem no 6.11,Page No.162¶

In :
import math

#Initilization of variables

L=2.5 #m #Length of beam
L_1=1.5 #m #Length from Fixed end

#Calculations

#Case-1
y=W*L**3*3**-1 #Deflection of the cantilever at free end

#Case-2
#Deflection of cantilever at free end is
#y_1=W_1*L**3*3**-1+W_1*L_1**3*3**-1+W_1*L_1**3*3**-1*(L-L_1)
#After substituting values in above equation and simplifying further we get

#y_1=22.375*W_1*3**-1

M_1=W*L*10**-3
M_2=W_1*L+W_1*L_1

#Let M_1=sigma_1*z  and M_2=sigma_2*z
#Dividing above two equations we get

#Let X=sigma_1*sigma_2**-1
X=M_2*M_1**-1*100

#Result
print"Max Bending stress is",round(X,2),"%"

Magnitude of equal Loads is 34.92 N
Max Bending stress is 111.73 %


## Problem no 6.12,Page No.163¶

In :
import math

#Initilization of variables

L=4 #m #Length of Beam

#calculations

#Consider a section at distance x from A and B.M at this section is
#M=P*(3-x)-10*x**2+90*x-195

#Now #EI*d**2*y*d**2*x=-P*(3-x)+10*x**2-90*x+195

#On Integrating above equation we get
#E*I*dy*dx**-1=-P*(3*x-x**2*2**-1)+10*x**3*2**-1-45*x**2+195*x+C_1

#Again On Integrating above equation we get
#E*I*y=-P*(3*x**2*2**-1-x**3*6**-1)+10*x**4*12**-1-15*x**3+195*x**2*2**-1+C_1*x+C_2

#But at x=0,dy*dx**-1=0 we get ,C_1=0
#        x=0,y=0     we get    ,C_2=0
#At x=3 m,y=0
x=3
C_1=0
C_2=0
P=(10*x**4*12**-1-15*x**3+195*x**2*2**-1+C_1*x+C_2)*(3*x**2*2**-1-x**3*6**-1)**-1

#Result

Load taken by prop is 60.0 KN


## Problem no 6.13,Page No.163¶

In :
import math

#Initilization of variables

L=6 #m #Span of Beam
sigma=100*10**6 #Pa #Bending stress
E=210*10**9
y=0.45 #m #Depth

#Calculations

#Taking moment at B
R_a=20*6*3+6*40*2*2**-1

#S.F=100-20*x-x**2*3**-1
#M=100*x-10*x**2-x**3*9**-1

#Thus B.M will be max where S.F is zero,we get equation as
#x**2+60*x-300=0
a=1
b=60
c=-300

X=b**2-4*a*c
x_1=(-b+X**0.5)*(2*a)**-1
x_2=(-b-X**0.5)*(2*a)**-1

x=4.641
M=100*x-10*x**2-x**3*9**-1 #KN*m #Max bending moment
I=M*sigma**-1*y*1000*2**-1 #m**4 #M.I

#E*I*d**2*y*(d*x**2)**-1=-100*x+10*x**2+x**3*9**-1

#AFter Integrating above EquATION WE get
#E*I*dy*(dx)**-1=-50*x**2+10*3**-1*x**3+x**4*36**-1+C_1
#Again Integrating above EquATION WE get
#E*I*y=-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1+C_1*x+C_2

#At x=0,y=0  ,C_2=0
#At x=6,y=0
x=6
C_2=0
C_1=-(-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1)*x**-1

x=3 #m
y=1*(E*I)**-1*(-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1+C_1*x+C_2)*1000*100

#Result
print"The central Deflection is",round(y,2),"m"

The central Deflection is 0.76 m


## Problem no 6.14,Page No.168¶

In :
import math

#Initilization of variables

L=10 #m #Lenght of cantilever beam
P_1=20*10**3 #N #Load at free end
P_2=20*10**3 #N #Load at middle of beam
E=200*10**9 #Pa
I=20000*10**-8 #m**4

#Calculations

#Taking moment at pt B we get
R_a=20*5*10**-1 #Force at pt A

#Now B.M at b=0,at C=-100,at A=-300 KN*m

#Now Area of B.M
A_1=2**-1*5*100 #KN*m**2
A_2=5*100       #KN*m**2
A_3=2**-1*5*200 #KN*m**2

#Total Area of B.M diagram is given by A
A=A_1+A_2+A_3

x_1=2*3**-1*5
x_2=3*2**-1*5
x_3=5*3**-1*5
M_1=A_1*x_1
M_2=A_2*x_2
M_3=A_3*x_3

M=M_1+M_2+M_3 #Total moments of B.M about B

y_B=M*10**3*(E*I)**-1 #Deflection a tfree end

#REsult
print"Slope of cantilever at free end is",round(theta,2),"radian"
print"Deflection of cantilever at free end is",round(y_B,2),"m"

Slope of cantilever at free end is 0.03 radian
Deflection of cantilever at free end is 0.22 m


## Problem no.6.15,Page no.169¶

In :
import math

#Initilization of variables

#Calculations

#Slope at A is Zero and deflection at C is zero According to Mohr's second theorem
#Let A_1*x_1=Y
Y=1*30**-1*80*4*(3*4**-1*4+2)
P=200*27**-1 #Reaction at ens D

#Result
print"The reaction at end C is",round(P,2),"KN"

The reaction at end C is 7.41 KN


## Problem no 6.16,Page No.171¶

In :
import math

#Initilization of variables

E=200*10**9 #Pa
I=2500*10**-8 #m**4

#Calculations

#Taking moment about A we get
R_a=(30*5+30*1)*6**-1 #Reaction at pt A
R_b=60-R_a #Reaction at pt B

M_c=30*1 #B.M at C
M_d=30*1 #B.M at D
M_a=0 #B.M at a
M_b=0 #B.M at b

#For conjugate beam taking moment about B_dash
R_a_dash=(30*2**-1*(5+1*3**-1)+30*4*3+30*2*3**-1*2**-1)*6**-1
R_b_dash=150-R_a_dash

y_e=1*(E*I)**-1*(R_a_dash*3-30*2*1-2**-1*1*30*(2+1*3**-1))*1000

#Result
print"Deflection at the centre is",round(y_e,2),"m"

Deflection at the centre is 0.03 m